Answer:
Part a)
Average EMF for half cycle is

Part b)
For one complete cycle we will have

Part c)
Maximum induced EMF will be at
t = 0.025 s and 0.075 s
minimum induced EMF is at
t = 0.05s and 0.1 s
Explanation:
As we know that magnetic field is oscillating in direction as well as magnitude
so induced EMF is given as

Part a)
For average value of EMF from positive maximum to negative maximum which is equal to half cycle
so we have




Part b)
For one complete cycle we will have


Part c)
Maximum induced EMF will be at

here we know

t = 0.025 s and 0.075 s
minimum induced EMF is at

so it is
t = 0.05s and 0.1 s
Answer:
v = R w
With this expression we see that for each point at different radius the tangential velocity is different
Explanation:
They indicate that the angular velocity is constant, that is
w = dθ / dt
Where θ is the radius swept angle and t the time taken.
The tangential velocity is linear or
v = dx / dt
Where x is the distance traveled in time (t)
In the definition of radians
θ = s / R
Where s is the arc traveled and R the radius vector from the pivot point, if the angle is small the arc (s) and the length (x) are almost equal
θ = x / R
We substitute in the speed equation
v = d (θ R) / dt
The radius is a constant for each point
v = R dθ / dt
v = R w
With this expression we see that for each point at different radius the tangential velocity is different
Answer:
Work is done by the heart on the blood during this time is 0.04 J
Explanation:
Given :
Mass of blood pumped, m = 80 g = 0.08 kg
Initial speed of the blood, u = 0 m/s
Final speed of the blood, v = 1 m/s
Initial kinetic energy of blood is determine by the relation:

Final kinetic energy of blood is determine by the relation:

Applying work-energy theorem,
Work done = Change in kinetic energy
W = E₂ - E₁

Substitute the suitable values in the above equation.

W = 0.04 J
Velocity is distance/time
so 150/7200=.0208km/s
unless you have to convert it to miles or something else. but use the formula!
Explanation:
Given that,
Mass if the rock, m = 1 kg
It is suspended from the tip of a horizontal meter stick at the 0-cm mark so that the meter stick barely balances like a seesaw when its fulcrum is at the 12.5-cm mark.
We need to find the mass of the meter stick. The force acting by the stone is
F = 1 × 9.8 = 9.8 N
Let W be the weight of the meter stick. If the net torque is zero on the stick then the stick does not move and it remains in equilibrium condition. So, taking torque about the pivot.

W = 3.266 N
The mass of the meters stick is :

So, the mass of the meter stick is 0.333 kg.